I encounter a problem in the proof below:

  • I don't know how to proove the first line in yellow (cf below): it makes me think about the Itô formula a lot

  • I don't undertand the deduction (ok $\gamma^{\chi}$ has the constant law of a brownian motion but what does it tell us ?)

Thank you

enter image description here

  • $\begingroup$ It is indeed confusing. Some notations, for example, $\varepsilon_u^{if} $, are used without definitions. $\endgroup$
    – Gordon
    May 11, 2015 at 15:39
  • $\begingroup$ Yes it is I 'll come back if I have the answer. I logically think that $\epsilon_{u}^{if}=e^{\int_{0}^{u}...}$ $\endgroup$
    – glork
    May 11, 2015 at 16:36

1 Answer 1


For the first question, since by definition, \begin{align*} \varepsilon_t^{if} = e^{i \int_0^{t}f\big(\frac{1}{\xi}\langle M\rangle_s\big)\frac{dM_s}{\sqrt{\xi}} + \frac{1}{2}\int_0^t f\big(\frac{1}{\xi}\langle M\rangle_s\big)\frac{d\langle M\rangle_s}{\xi}}, \end{align*} then, \begin{align*} d\varepsilon_t^{if} = i \varepsilon_t^{if} f\Big(\frac{1}{\xi}\langle M\rangle_t\Big)\frac{dM_t}{\sqrt{\xi}}. \end{align*} Moreover, \begin{align*} \langle \varepsilon_t^{if}, H_t \rangle = \int_0^t i \varepsilon_s^{if} f\Big(\frac{1}{\xi}\langle M\rangle_s\Big) h_s \frac{d\langle M\rangle_s}{\sqrt{\xi}}, \end{align*} as $\langle M_t, R_t \rangle = 0$. Consequently, \begin{align*} d\big(\varepsilon_t^{if} H_t\big) &= H_t d\varepsilon_t^{if} + \varepsilon_t^{if} dH_t + d \langle \varepsilon_t^{if}, H_t \rangle\\ &= H_t d\varepsilon_t^{if} + \varepsilon_t^{if} dH_t + i \varepsilon_t^{if} f\Big(\frac{1}{\xi}\langle M\rangle_t\Big) h_t \frac{d\langle M\rangle_t}{\sqrt{\xi}}. \end{align*} That is, \begin{align*} \mathbb{E}\big(\varepsilon_{\infty}^{if} H_{\infty}\big) -\mathbb{E}(H_0) &= \mathbb{E}\bigg(\int_0^{\infty}\!\!\!\! H_t d\varepsilon_t^{if} + \int_0^{\infty} \!\!\!\!\varepsilon_t^{if} dH_t + i\int_0^{\infty}\!\!\!\!\varepsilon_t^{if} f\Big(\frac{1}{\xi}\langle M\rangle_t\Big) h_t \frac{d\langle M\rangle_t}{\sqrt{\xi}} \bigg) \\ &=i \mathbb{E}\bigg(\int_0^{\infty}\!\!\!\!\varepsilon_t^{if} f\Big(\frac{1}{\xi}\langle M\rangle_t\Big) h_t \frac{d\langle M\rangle_t}{\sqrt{\xi}} \bigg), \end{align*} or \begin{align*} \mathbb{E}\big(\varepsilon_{\infty}^{if} H\big) = \mathbb{E}(H)+i \mathbb{E}\bigg(\int_0^{\infty}\!\!\!\!\varepsilon_t^{if} f\Big(\frac{1}{\xi}\langle M\rangle_t\Big) h_t \frac{d\langle M\rangle_t}{\sqrt{\xi}} \bigg). \end{align*}


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